Question

Prove that a cyclic trapezium is always an isosceles trapezium

Answer 1

et ABCD be the cyclic trapezium with AB IICD.

thru' C draw CE parallel to  AD meeting AB in E.

So

AECD is a parallelogram.

so

angle D=angle AEC.... opp angles of a parallelogram are equal....(i)

but

angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii)

from (i) and (ii)

angle AEC+angle ABC=180

but

angle AEC+angle CEB= 180...linear pair

so

 angle ABC= angle CEB ..(iii)

so

 CE=CB...  sides opp equal angles are equal.(iv)

 but

CE=AD...opp  sides of parallelogram AECD.

from (iv) we get

AD=CB

Thus cyclic quadri ABCD is isoceles.

this proves the first part of the question.

now,

 join AC and BD, the diagonals.

in triangles DAB and CBA,

AD=CB...proved before

AB=AB common

angle ADB= angle ACB.. angles in the same  segment of a circle are equal.here AB is the chord.

so triangles DAB and CBA are congruent....SAS rule.

so

AD=CB... CPCT

hence proved.

 

Answer 2

Given : cyclic trapezium is an isosceles trapezium

To Fine : prove

Solution:

Let say ABCD is a cyclic trapezium  where

AD || BC

Hence we need to show that AB = CD for it to be an isosceles trapezium

as its  cyclic => sum of opposite angles = 180°

Lets extend AD  and draw a line parallel to AB passing through C and intersecting  extended AD at  E.

AB || CD  and  CE || AB

hence ABCE is a parallelogram

=> AB = CE  opposite sides are equal

in parallelogram opposite angles are equal

Hence ∠B = ∠E

∠B + ∠D  = 180°    ( cyclic Quadrilateral )

∠D + ∠CDE  = 180°   linear Pair

=> ∠E = ∠CDE

in ΔCDE

∠E = ∠CDE

=> CD = CE

  AB = CE    ( opposite sides of parallelogram )

=> AB = CD

Hence trapezium is isosceles trapezium

QED

Hence proved

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